Question

If $\frac{n!}{2!(n-2)!}$ and $\frac{n!}{4!(n-4)!}$ are in the ratio $2:1$, find the non-zero value of $n$.

Answer 1

Given, $\frac{n!}{2!(n-2)!}:\frac{n!}{4!(n-4)!}=2:1$
$\Rightarrow \frac{n!}{2!(n-2)!}\times \frac{4!(n-4)!}{n!}=\frac{2}{1}$
$\Rightarrow \frac{4!(n-4)!}{2!(n-2)}=2$
$\Rightarrow \frac{4\times 3\times 2!}{2!}\times \frac{4!(n-4)!}{(n-2)(n-3)(n-4)!}=2$
$\Rightarrow 12=2(n-2)(n-3)$
$\Rightarrow 6=n^2-5n+6$
$\Rightarrow n^2-5n=0$
$\Rightarrow n(n-5)=0$
$\Rightarrow n=0$ or $5$